Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three vriables, where $\delta_d = \frac{d(d+3)}{2}$. Let $$ X \subset \mathcal{D} \times \mathbb{P}^2$$

be a smooth embedded complex submanifold, not necessarily closed. Given a point $p\in \mathbb{P}^2$, we get a hyperplane $$\tilde{H}_p \in \mathcal{D} \times \mathbb{P}^2.$$ Note that a point $p$ first of all gives a hyperplane $H_p$ in $\mathcal{D}$ (which is the space of degree $d$ polynomials passing through the point $p$). This gives us a hyperplane $$ \tilde{H}_p := H_p\times \mathbb{P}^2 \in \mathcal{D} \times \mathbb{P}^2.$$ Is it true that if $\tilde{H}_p$ is not transverse to $X$, then for every open neighborhood $U$ of $p$ in $\mathbb{P}^2$, there exists a point $q\in U$ such that the hyperplane $\tilde{H}_q$ does intersect $X$ transversally? This seems to intuitively say that if something is not transverse then you can ``perturb'' it so that it is transverse. More precisely, I want to claim that the set of points $q$ where $\tilde{H}_q$ intersects $X$ transversally is an open, dense subset of $\mathbb{P}^2$. It is certainly open. Please note that apriori it is possible that there does not exist any point $q$ such that $\tilde{H}_q$ intersects $X$ transversally. This is the part I don't see how to prove, although intuitively it seems obvious.